AMC 10 · 2007 · #8
Grade 8 geometry-2dTriangles ABC and ADC are isosceles with AB=BC and AD=DC. Point D is inside triangle ABC, angle ABC measures 40 degrees, and angle ADC measures 140 degrees. What is the degree measure of angle BAD?
Pick an answer.
AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two isosceles triangles share the base $AC$: triangle $ABC$ has $AB=BC$ with apex angle $\angle ABC=40^\circ$, and triangle $ADC$ has $AD=DC$ with apex angle $\angle ADC=140^\circ$. Point $D$ sits inside triangle $ABC$. Find the measure of $\angle BAD$.
Givens: Triangle $ABC$ is isosceles with $AB=BC$; Triangle $ADC$ is isosceles with $AD=DC$; $\angle ABC=40^\circ$ and $\angle ADC=140^\circ$; $D$ is a point inside triangle $ABC$; Answer choices: (A) $20$, (B) $30$, (C) $40$, (D) $50$, (E) $60$ degrees
Unknowns: The measure, in degrees, of $\angle BAD$
Understand
Restated: Two isosceles triangles share the base $AC$: triangle $ABC$ has $AB=BC$ with apex angle $\angle ABC=40^\circ$, and triangle $ADC$ has $AD=DC$ with apex angle $\angle ADC=140^\circ$. Point $D$ sits inside triangle $ABC$. Find the measure of $\angle BAD$.
Givens: Triangle $ABC$ is isosceles with $AB=BC$; Triangle $ADC$ is isosceles with $AD=DC$; $\angle ABC=40^\circ$ and $\angle ADC=140^\circ$; $D$ is a point inside triangle $ABC$; Answer choices: (A) $20$, (B) $30$, (C) $40$, (D) $50$, (E) $60$ degrees
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Nothing here needs algebra — the whole problem is reading angles off a picture, so Tool #1 (Draw a Diagram) comes first: sketch both isosceles triangles on the shared base $AC$ and place $D$ inside triangle $ABC$. The picture reveals the key relationship — ray $AD$ falls between rays $AB$ and $AC$, so $\angle BAC$ splits into $\angle BAD$ and $\angle DAC$. That turns the goal into 'find $\angle BAC$ and $\angle DAC$ separately, then subtract,' which is Tool #7 (Identify Subproblems). Each of those two angles is a base angle of one isosceles triangle, found from the triangle's angle sum.
Execute — Answer: D
4.MD.C.7 Step 1 Draw the figure and split the angle at A
- Sketch segment $AC$ and build both triangles on it: apex $B$ above with $AB=BC$, and apex $D$ inside triangle $ABC$ with $AD=DC$.
- Look at vertex $A$.
- Ray $AB$ goes to the outer apex, ray $AC$ runs along the base, and because $D$ is inside the triangle, ray $AD$ lies between them.
- So the angle $\angle BAC$ is cut into two pieces: $\angle BAC=\angle BAD+\angle DAC$.
- Rearranged, $\angle BAD=\angle BAC-\angle DAC$.
- Now the job is to find those two base angles.
💡 Since $D$ sits inside the triangle, ray $AD$ splits $\angle BAC$, so the target angle is just the big base angle minus the small one.
8.G.A.5 Step 2 Find angle BAC from triangle ABC
- In triangle $ABC$ the apex angle is $\angle ABC=40^\circ$, and the three angles sum to $180^\circ$, leaving $180^\circ-40^\circ=140^\circ$ for the two base angles $\angle BAC$ and $\angle BCA$.
- Because $AB=BC$, those two base angles are equal, so each is half of $140^\circ$: $\angle BAC=\dfrac{180^\circ-40^\circ}{2}=70^\circ$.
💡 Equal sides face equal angles, so the leftover after the apex splits evenly between the two base angles.
8.G.A.5 Step 3 Find angle DAC from triangle ADC
- Do the same for triangle $ADC$.
- Its apex angle is $\angle ADC=140^\circ$, so the two base angles $\angle DAC$ and $\angle DCA$ share what is left: $180^\circ-140^\circ=40^\circ$.
- Since $AD=DC$, these base angles are equal, so each is half of $40^\circ$: $\angle DAC=\dfrac{180^\circ-140^\circ}{2}=20^\circ$.
💡 A wide apex angle of $140^\circ$ leaves only a little for the base angles, so each is a small $20^\circ$.
4.MD.C.7 Step 4 Subtract to get angle BAD
- Put the two base angles into the relationship from the picture: $\angle BAD=\angle BAC-\angle DAC=70^\circ-20^\circ=50^\circ$.
- So $\angle BAD=50^\circ$, which is choice (D).
- The other options come from slips: (C) $40^\circ$ just echoes the given apex angle, (A) $20^\circ$ stops at $\angle DAC$, and (B) $30^\circ$ or (E) $60^\circ$ come from halving or combining the wrong angles.
💡 With both base angles known, the answer is a single subtraction of the small angle from the big one.
4.MD.C.7 Sketch segment $AC$ and build both triangles on it: apex $B$ above with $AB=BC$, 8.G.A.5 In triangle $ABC$ the apex angle is $\angle ABC=40^\circ$, and the three angles 8.G.A.5 Do the same for triangle $ADC$. Its apex angle is $\angle ADC=140^\circ$, so the 4.MD.C.7 Put the two base angles into the relationship from the picture: $\angle BAD=\ang Review
Reasonableness: The answer must be a positive angle smaller than $\angle BAC=70^\circ$, since $\angle DAC=20^\circ$ is carved out of it — and $50^\circ$ fits. Sanity-check the whole picture: at $A$ the pieces $20^\circ+50^\circ=70^\circ=\angle BAC$; triangle $ABC$ has $70^\circ+70^\circ+40^\circ=180^\circ$ and triangle $ADC$ has $20^\circ+20^\circ+140^\circ=180^\circ$. Everything closes up, so $50^\circ$ is consistent.
Alternative: Use the symmetry instead of the split at $A$. Both apexes $B$ and $D$ lie on the perpendicular bisector of $AC$ (each isosceles triangle is symmetric across it), so reflecting across that line swaps $A$ and $C$ while fixing $B$ and $D$ — hence $\angle BAD=\angle BCD$. Working on the $C$ side, $\angle BCA=70^\circ$ and $\angle DCA=20^\circ$ (the equal base angles), so $\angle BCD=\angle BCA-\angle DCA=70^\circ-20^\circ=50^\circ$. This independently gives $\angle BAD=50^\circ$.
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about the angle sum of triangles (Getting each base angle from its triangle's angle sum: $\angle BAC=70^\circ$ from the $40^\circ$ apex and $\angle DAC=20^\circ$ from the $140^\circ$ apex, using that equal sides give equal base angles.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems to find unknown angles (Splitting $\angle BAC$ into $\angle BAD+\angle DAC$ because $D$ is inside the triangle, then subtracting to get $\angle BAD=70^\circ-20^\circ=50^\circ$.)
⭐ In an isosceles triangle the two base angles are equal, so split the leftover after the top angle in half; here that gives $70^\circ$ and $20^\circ$, and their difference is the answer.
⭐ In an isosceles triangle the two base angles are equal, so split the leftover after the top angle in half; here that gives $70^\circ$ and $20^\circ$, and their difference is the answer.
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