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Areas Of Parallelograms And Triangles

Question
CBSEENMA9002583

ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see figure). Show that these altitudes are equal.


Solution

Given: ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively.
To Prove: BE = CF.
Proof: ∵ ABC is an isosceles triangle
∴ AB = AC
∴ ∠ABC = ∠ACB    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆BEC and ∆CFB,
∠BEC = ∠CFB    | Each = 90°
BC = CB    | Common
∠ECB = ∠FBC    | From (1)
∴ ∆BEC ≅ ∆CFB    | By AAS Rule
∴ BE = CF.    | C.P.C.T.