Question

in isosceles triangle and,AB,AC.Perpendicular bd and ce are drawn from the vertices b and c,to the opposite sides.show that bd =ce​

Answer 1

In an isosceles triangle ABC, the perpendiculars BD and CE to the sides of triangle AC and AB are equal to each other.

• Given,
• ABC is an isosceles triangle
• Since, we know that, in an isosceles triangle, two sides  are of equal length and the opposites angles are same, we get,

• AB=AC
• and
• ∠ABC=∠ACB ......(a)
• ⇒$\frac{1}{2}AB=\frac{1}{2}AC$

• Since BD and CE are the medians of triangle, we have
• BE=CD......(b)
• In ΔBCD and ΔCEB
• from (a)
• ∴∠BCD=∠CBE

• and
• from (b)
• BE=CD
• BC=CB   (common)
• ∴ΔBCD≅ΔCEB    (SAS congruence rule)
• ∴BD=CE

Answer 2

BD = CE. (Proved)

Step-by-step explanation:

We have an isosceles triangle Δ ABC with equal sides AB and AC.

Now, CE is the perpendicular on AB from vertex C and BD is the perpendicular on AC from vertex B.

So, the area of the triangle Δ ABC will be given by

Area = $\frac{1}{2}\times AB \times CE = \frac{1}{2}\times AC \times BD$

Now, since, AB = AC, then we can say that CE = BD or BD = CE. (Proved)