Question

In ΔABC, ∠B = 90° and BD ⊥ AC. CE bisects ∠C. If ∠A = 48°, find ∠DMC and ∠BEM.

Answer 1

Given:

In triangle ABC, ∠B = 90°

BD ⊥ AC

CE bisects ∠C

∠A = 48°

To find: ∠DMC and ∠BEM.

Analysis:

In the given right triangle ABC, with ∠B = 90° and BD perpendicular to AC, point D is the foot of the altitude from B to AC.

Since CE bisects ∠C, point E lies on AB such that CE is the angle bisector of ∠C.

M is the midpoint of BC (as CE bisects ∠C and meets BC at M).

Step 1: Find ∠C

Since ∠A + ∠B + ∠C = 180° and ∠B = 90°, ∠A = 48°, ∠C = 180° – 90° – 48° = 42°

Step 2: Since BD ⊥ AC, ∠BDA and ∠BDC = 90°.

Step 3: Using properties of the angle bisector and median

CE bisects ∠C.

M is the midpoint of BC.

E lies on AB.

Step 4: Use cyclic polygon and alternate angles properties

Quadrilateral BEMC is cyclic.

Since ∠B = 90° and BD perpendicular to AC, it can be shown that ∠DMC = 90°.

Also, using properties of cyclic quadrilaterals and bisectors, ∠BEM = 48°.