You visited us 1 times! Enjoying our articles? Unlock Full Access!

Similarity of Triangles

In the given ...

Question

In the given triangle ABC, altitudes BD and CE to sides AC and AB are equal. Show that ΔABD ≅ ΔACE.

Open in App

Solution

Given that BD ⊥ AC; CE ⊥ AC
BD = CE
Now in ΔABD and ΔACE
∠ADB = ∠AEC (∵ given 90°)
∠A = ∠A (common angle)
BD = CE
∴ ΔABD = ΔACE (∵ AAS congruence)
⇒ AB = AC (∵ C.P.C.T)

Suggest Corrections

Similar questions

A B C

B D

C E

A C

A B

△ A B D ≅ △ A C E

ABC

BD

CE

AC

AB

AB = AC

ABC

A B C

B D

C E

A C

A B

$A B C$ is a triangle in which altitudes $B E$ and $C F$ to sides $A C$ and $A B$ are equal (see the given figure). Show that

(i) $△ A B E ≅ △ A C F$

(ii) $A B = A C,$ i.e., $A B C$ is an isosceles triangle.

Fill in the blanks in the following so that each of the following statements is true.

(i) Sides opposite to equal angles of a triangle are ___

(ii) Angle opposite to equal sides of a triangle are ___

(iii) In an equilateral triangle all angles are ___

(iv) In a $Δ A B C i f ∠ A = ∠ C, t h e n A B =$ ___

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ___

(vi) In an isosceles, triangle ABC with AB = AC, if BD and CE are its altitudes then BD is ___ CE.

(vii) In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then $Δ A B C ≅ Δ$ ___

Join BYJU'S Learning Program