Question

In ∆ABC, altitude AD bisects BC, prove that ∆ADB ≅ ∆ADC. Write equal pairs of sides and angles of these two triangles.

c291ee94aed4cd51dec5b13d35496986.pptx

Answer 1

In tri. ADB & ADC seg AD = seg AD …(Common Side) Side BD = side DC …(Altitude bisects side BC) tri. ADB = ADC is congruent …(By Hypotenuse side theorem) Side AB = Side AC …(c.s.c.t) Angle B = Angle C …(c.a.c.t)

Answer 2

The pairs of sides and angles of these two triangles are. AB ≅ AC, ∠ABD ≅ ∠ACD, ∠BAD ≅ ∠CAD.        

Step-by-step explanation:

Given:

In ∆ABC,

altitude AD bisects BC,

To Prove:

∆ADB ≅ ∆ADC.

Proof:

In  Δ ADB and Δ ADC

AD ≅ AD                        ....……….{ Common Side}

∠ADB ≅ ∠ADC = 90°    …………..{Altitudes are Perpendicular}

BD ≅ CD                        ....……….{AD bisects BC}

Δ ADB ≅ Δ ADC           ….{Side-Angle-Side test}

∴ AB ≅ AC                    ....{corresponding parts of congruent Triangle }

∴ ∠ABD ≅ ∠ACD         .....{corresponding parts of congruent Triangle }

∴ ∠BAD ≅ ∠CAD          ....{corresponding parts of congruent Triangle }