Question

In ∆ABC, (a) If (b) Prove that angles of ∆ACD and ∆BCD are equal.
(c) If the perpendicular from the right angles vertex of a right triangle divides the
opposite side into parts of lengths a and b and if the length of the perpendicular is
h, then prove that h
2 = ab.

Answer 1

To Prove : △ADE∼△ACB

Proof :

(i) In △ADE and △ACB

   (1) ∠A=∠A                [common]

   (2) ∠AED=∠ABC=90o  (given)

   ∴    △ADE∼△ACB  [AA axiom]

(ii) (AC)2=(AB)2+(BC)2

    169=(AB)2+25

    AB=12cm

∵   △ADE∼△ACB

∴   BCDE​=ACAD​=ABAE​

∴     BCDE​=ABAE​

             5DE​=124​

             DE=1220​=35​=132​cm

Now, ACAD​=ABAE​

        13AD​=124​

        AD=1213×4​=313​=431​cm.

(iii) Ar.of(△ADE)Ar.of(△ABC)​=AE2AB2​=16144​=19​

   Ar.of(△ADE)Ar.of(△ADE)+Ar.of(BCED)​=9

   1+Ar.of(△ADE)Ar.of(BCED)​=9

   Ar.of(BCED)Ar.of(△ADE)​=81

please mark as brainliest