In the fig- given below- BD and CE are two altitudes of - ABC- If altitude BD - CE- then - ABC is an - triangle-

In Δ ABC and ACE, we have ∠ADB = ∠AEC = 90^∘ ∠BAD = ∠CAE [Common angle] and, BD = CE [Given] So, by AAS criterion of congruence, we ...

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RHS Criteria for Congruency

In the fig. g...

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In the fig. given below, BD and CE are two altitudes of $Δ$ ABC. If altitude BD = CE, then $Δ$ ABC is an __ triangle.

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In Fig. BD and CE are two altitudes of a $Δ$ ABC such that BD = CE.

Prove that the ABC is an isoceles triangle. [2 MARKS]

. . . .

Δ A B C

Δ

A B = A C

B D

C E

B D = C E

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 $Δ A B C ≅ Δ$ ___

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