Question

Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC.

Answer 1

)) AB = AC(Given)

∴ ∠FBE = ∠ACD

∠BFE = ∠ADC

ΔEFB ~ ΔADC (AA similarity


(ii) Similarly, it can be proved that ΔADB ~ ΔEFB

Answer 2

(i) AB = AC(Given)

∴ ∠FBE = ∠ACD

∠BFE = ∠ADC

ΔEFB ~ ΔADC (AA similarity)

$\frac{ar(adc)}{ar(efb)} = {( \frac{ac}{be} )}^{2} \\ = {( \frac{ac}{bc + ce} )}^{2} \\ = {( \frac{13}{18} )}^{2} \\ = \frac{169}{329}$


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