Question

In triangle ABC , AB = AC . If AD is the median from A , prove that it is an altitude also .​

Answer 1

Answer:

Given: ABC is an isosceles triangle(AB=AC)

           AD is the altitude

To prove: AD is the median

Proof:

AB=AC (given)

AD is common

∠ADB=∠ADC=90° (AD is the height)

∴By RHS criteria, ΔABD is congruent to ΔACD

By CPCT,

BD=CD

And area of congruent triangles are equal.

A median from a vertex divides the opposite side into two halves and the two triangles formed are equal in area.

Hence AD is the median as well as height.

Step-by-step explanation:

Answer 2

✯ ᴀɴsᴡᴇʀ ✯

★ given →

☞︎︎︎ in triangle ABC , AB = AC and AD is the median of triangle

_____________________

★ to prove →

☞︎︎︎ AD is altitude of triangle ABC

_____________________

★ proof →

☞︎︎︎ In triangle ABD and ACD

・AB = AC (given)

・BD = CD (since AD is median of ∆ABC)

・AD = AD (common)

➪ ∆ABD ≅ ∆ACD by SSS congruency rule

now, ∠BDA = ∠CDA (by C.P.C.T) ........(i)

ᗒ C.P.C.T = congruent parts of congruent triangles

we know that, BDC is a straight line

so, ∠BDA + ∠CDA = 180° ( by linear pair or sum of angles of a straight line is 180°)

➪∠BDA + ∠CDA = 180°

form eq.(i) ∠BDA = ∠CDA

➪ ∠BDA + ∠BDA = 180°

➪ 2∠BDA = 180°

➪ ∠BDA = 180°/2

➪ ∠BDA = 90°

➪ ∠BDA = ∠CDA = 90°

since ∠BDA and ∠CDA are right angles,

so we can say that AD is perpendicular to BC

and we know that, perpendicular line of a triangle is altitude of that triangle

so, that AD is median and altitude of triangle ABC

hence, proved!