IPMAT Indore 2024 · Quantitative Ability · Question 20

Step 1 — name the base angles. Since AB=AC, let \angle ABC=\angle ACB=\beta. Then \angle BAC={180}^{∘}-2\beta.

Step 2 — split the apex. Given \angle BAD={30}^{∘},

\angle DAC=\angle BAC-\angle BAD=\left({180}^{∘}-2\beta \right)-{30}^{∘}={150}^{∘}-2\beta

Step 3 — find \angle ADC from triangle ADC.

\angle ADC={180}^{∘}-\angle DAC-\angle ACD={180}^{∘}-\left({150}^{∘}-2\beta \right)-\beta ={30}^{∘}+\beta

Step 4 — use AD=AE. Triangle ADE is isosceles with apex angle \angle DAE=\angle DAC={150}^{∘}-2\beta, so its base angles are

\angle ADE=\angle AED=\frac{{180}^{∘}-\left({150}^{∘}-2\beta \right)}{2}=\frac{{30}^{∘}+2\beta}{2}={15}^{∘}+\beta

Step 5 — subtract.

\angle CDE=\angle ADC-\angle ADE=\left({30}^{∘}+\beta \right)-\left({15}^{∘}+\beta \right)={15}^{∘}

The \beta cancels — the answer does not depend on the shape of the isosceles triangle at all.

Answer: Option 1 — {15}^{∘}

Shortcut. Learn the result: \angle CDE=1/2\angle BAD. Whenever AB=AC, D on BC, E on AC with AD=AE, the angle \angle CDE is always half of \angle BAD, independent of the triangle’s shape.

Since the answer is shape-independent, you may simply pick a convenient triangle — take \beta ={60}^{∘} (equilateral), so \angle BAC={60}^{∘}, \angle DAC={30}^{∘}, \angle ADC={90}^{∘}, \angle ADE={75}^{∘}, and \angle CDE={15}^{∘}. Thirty seconds, no algebra.