Question

AP is altitude of an isosceles traingle with AB=AC. Show that AngleB =AngleC.​

Answer 1

$\large\underline{\sf{Given- }}$

An isosceles triangle ABC such that,

AB = AC

AP is altitude.

$\large\underline{\sf{To\:prove - }}$

∠B = ∠C

$\large\underline{\sf{Solution-}}$

Given that,

An isosceles triangle ABC such that AB = AC

and

AP is altitude.

⟹ ∠APB = ∠APC = 90°

Now, Consider,

$\red{\rm :\longmapsto\:\triangle \: APB \: and \: \triangle \: APC}$

$\rm :\longmapsto\:AB \: = \: AC \: \: \: \: \: \: \: \{given \}$

$\rm :\longmapsto\:AP \: = \: AP \: \: \: \: \: \: \: \{common \}$

$\rm :\longmapsto\:\angle \: APB \: = \: \angle \: APC \: \: \: \: \: \: \: \{each \: 90 \degree \}$

$\red{\rm :\longmapsto\:\triangle \: APB \: \cong \: \triangle \: APC \: \: \: \: \{RHS \: Rule \}}$

$\bf\implies \:\:\angle \: B \: = \: \angle \: C \: \: \: \: \: \: \: \{CPCT \}$

Hence, Proved

Additional Information  :

1) AAS Congruency  :

If two angles and one side of one triangle is equal to two angles and one side of a triangle, then they are congruent.

Example :

In ΔABC and ΔDEF, ∠A = ∠D, ∠B = ∠E and BC= EF then ΔABC ≅ ΔDEF by AAS criteria.

2) ASA Congruency  :

If two angles and included side of one triangle are respectively equal to two angles and included side of another triangle, then the two triangles are congruent.

Example :

In ΔABC and ΔDEF, ∠A = ∠D, ∠C = ∠F and AC = DF then ΔABC ≅ ΔDEF by ASA criteria.

3) SSS Congruency :

If the corresponding sides of two triangles are equall, then the two triangles are congruent.

Example :

In ΔXYZ and ΔLMN, XY = LM, YZ = MN and XZ = LN then ΔXYZ ≅ ΔLMN by SSS criteria.

4) SAS Congruency :

If in two triangles, one pair of corresponding sides are equall and the included angles are equal then the two triangles are congruent.

Example :

In ∆ABC & ∆DEF,∠A = ∠D, AB = DE, AC = DF then ∆ABC ≅ ∆DEF by SAS criteria.

5) RHS Congruency

If in two triangles, right angle, Hypotenuse and one side are equal, then triangles are congruent.

Example :

In ∆ABC & ∆DEF,∠A = ∠D = 90°, AB = DE, BC = EF then ∆ABC ≅ ∆DEF by RHS criteria.