Question

Suppose ABC is a triangle and D is the midpoint of BC. Assume that the perpendiculars from D to AB and AC are of equal length. Prove that ABC is isosceles.

Answer 1

Given: PD = DQ, BD = DC and ∠DPB = ∠DPA = ∠DQC = ∠DQA = 90°

To Prove: AB = AC

Proof:

In ΔAPD and ΔAQD:

PD = DQ (Given)

∠APD = ∠AQD = 90° (Given)

AD = AD (Common)

∴ ΔAPD ≅ ΔAQD (RHS congruency)

⇒ AP = AQ … (1) (Corresponding parts of congruent triangles)

In ΔBPD and ΔCQD:

PD = DQ (Given)

∠BPD = ∠CQD = 90° (Given)

BD = DC (Given)

∴ ΔBPD ≅ ΔCQD (RHS congruency)

⇒ BP = QC … (2) (Corresponding parts of congruent triangles)

From (1) and (2):

AP = AQ and BP = QC

AP + BP= AQ +QC

∴ AB = AC

Thus, the given triangle is isosceles.