Question

BE and CF are two equal altitudes of a triangle ABC . Using RHS congruence rule , prove that the triangle ABC is isosceles .​

Answer 1

Step-by-step explanation:

Given:

• BE and CF are two altitudes of triangle ABC.
• BE is also equal to CF.

To Prove:

• ∆ABC is an isosceles triangle i.e AB = AC

Proof: In ∆BCF and ∆CBE , we have

• ∠BFC = ∠CEB = 90°

• BC = CB ( Common in both )

• FC = EB ( Given , altitudes are equal )

So, ∆BCF ≅ ∆CBE , by RHS congruence rule.

∴ ∠FBC = ∠ECB ( By C.P.CT )

As we know that if two angles are equal then the sides opposite to that angles are also equal.

=> we got AB = AC

Hence, ∆ABC is an isosceles triangle.

Answer 2

$\sf{\underline{\underline{Given:}}}$

$\sf{Given,\:BE\:is\:a\:altitude,\:So,}$

$\sf\red{\angle AEB = \angle CEB = 90\degree \: \: \: \: \: ......(1)}$

$\sf{\underline{\underline{Also:}}}$

$\sf{CF \:is\:altitude,So,}$

$\sf\green{\angle AFC = \angle BFC = 90\degree\: \: \: \: \: .....(2)}$

$\sf\orange{Also,\: BE = CF \: \: \: \: \: ......(3)}$

$\sf\large{\underline{\underline{To\:Prove:}}}$

$\sf{Triangle\:ABC\:is\:Isosceles}$

$\sf\large{\underline{\underline{Proof:}}}$

$\sf{\longrightarrow In\:Triangle\:BCF \:and\:Triangle\:CBE}$

$\sf{\red{ \angle BCF = \angle CEB = 90\degree \: \: \: ...(Both\:90\degree)}}$

$\sf{\purple{BC = CB \: \: \: .....(Common)}}$

$\sf{\green{FC = EB \: \: \: ...(From (3))}}$

$\sf\orange{Triangle\:BCF\:is\:congruence\:to\: Triangle\:CBE \: \: \: ....(RHS\:rule\:of\:Congruency)}$

$\sf\large{\underline{\underline{Therefore:}}}$

$\sf\red{\angle FBC = \angle ECB \: \: \: \: \: .....(CPCT)}$

$\sf\green{So,\: \angle ABC = \angle ACB}$

$\sf\purple{ AB = AC \: \: \: ...(Sides\:opposite\:to\:equal\:angles\:is\:equal)}$

$\sf\large{\underline{\underline{Hence:}}}$

$\bold{\fbox{\color{red}{Triangle\:ABC\:is\:an\:Isosceles\:Triangle.}}}$