Question

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS
rule, prove that the triangle ABC is isosceles.
5. ABC is an isosceles triangle with AB = AC. Draw AP 1 BC to sho
ZB= Z C.
side
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Answer 1

$\huge{\underline{\tt{\red{Answer-}}}}$

Consider triangle BFC AND CEB

BC = CB. ( COMMON )

ANGLE BFC = ANGLE CEB = 90 (GIVEN)

CF = BE. ( GIVEN )

SO, triangle BFC is congruent to triangle CEB. ( RHS )

$Hence , BF = CE. ( CPCT )---(1)$

and, considered triangle ABE and ACF

angle BAC = angle CAB. ( COMMON )

angle AFC = angle AEB. (GIVEN)

CF = BE. (GIVEN)

HENCE,. triangle ABE is congruent to triangle ACF. (AAS)

$So, AF = AE. (CPCT). ---(2)$

adding (1) and (2)

we get,

BF + AF = CE + AE

AB = AC

hence , triangle ABC is iso. triangle.

Answer 2

$\huge\underline\mathbb{TRIANGLES}$

In $\triangle$ BFC and $\triangle$ BEC

=> BC = BC (Given)

=> BE = CF (Given)

=> $\angle$ BEC = $\angle$ CFB ( Each 90°)

$\therefore$ Triangle BFC is congurent to triangle BEC ( RHS RULE)

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$\angle$ B = $\angle$C (CPCT)

So,

AB = AC ( sides opposite to equal angles of a triangle are equal)

$\therefore$ ABC is an isosceles triangle.