In a right-angled ΔABC, the perpendicular BD on hypotenuse AC is drawn. Prove that :
(i) AC × AD = AB²
(ii) AC × CD = BC²

Question

In a right-angled ΔABC, the perpendicular BD on hypotenuse AC is drawn. Prove that : (i) AC × AD = AB2 (ii) AC × CD = BC2

KnowledgeBoat · Accepted Answer

(i) In ΔABC and ΔADB,

⇒ ∠ABC = ∠ADB = 90°

⇒ ∠BAC = ∠DAB [Common angles]

△ABC ∼ △ADB [By AA similarity]

Since, corresponding sides of similar triangle are proportional to each other.

$\dfrac{AB}{AD} = \dfrac{AC}{AB}$

AB² = AC × AD

Hence, AB² = AC × AD.

(ii) In ΔABC and ΔBDC

⇒ ∠ABC = ∠CDB = 90°

⇒ ∠C [Common angles]

△ABC ∼ △BDC [By AA similarity]

$\dfrac{BC}{CD} = \dfrac{AC}{BC}$

BC² = AC × CD

Hence, BC² = AC × CD.

Mathematics