Question

AB =10 BC=15 DE perpendicular to AC AD angel bisector of A then , find DE​

Answer 1

Step-by-step explanation:

Given :

∠A = 90°

AD is the bisector of ∠A

DE ⊥ AC

AB = 10 cm

AC = 15 cm

To find :

The length of DE

Concept used :

Similarity of Triangle

As AD is the bisector of ∠CAB

So, ∠EAD = 90°/2

⇒ ∠EAD = 45°

In ΔEAD

⇒ ∠AED + ∠EAD + ∠EDA = 180°

⇒ 90° + 45° + ∠EDA = 180°

⇒ ∠EDA = 180° - 135°

⇒ ∠EAD = 45°

So. EA = ED

Let ED be x

⇒ EC = 15 - x

In ΔCED and ΔCAB

⇒ ∠C = ∠C (common)

⇒ ∠CED = ∠CAB = 90°

So, ΔCED ~ ΔCAB (AA)

⇒ (ED)/(AB) = (CE)/(CA)

⇒ x/10 = (15 - x)/15

⇒ 15x = 10(15 - x)

⇒ 15x = 150 - 10x

⇒ 15x + 10x = 150

⇒ 25x = 150

⇒ x = 150/25

⇒ x = 6

∴ The length of DE is 6 cm.