Question

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Answer 1

Answer:

I am considering the triangle ABC as an equilateral triangle, as you haven't given any specifics in your question.

Step-by-step explanation:

SOLUTION :

Given

ABC is a triangle in which BE⊥ AC and BE and CF intersect each other at O.

TO PROVE

AD ⊥ BC.

CONSTRUCTION

Join EF .

Since base BC makes equal angles at E and F (each equal to 90°), therefore, these points must lies on circumference of a circle.

∴ BCEF is a cyclic quadrilateral.

∴ ∠CBF + ∠CEF = 180°

⟹ ∠CBF + ∠CEB + ∠BEF = 180°

But, ∠CEB = 90° [∴BE⊥ AC]

So, ∠CBF + 90° + ∠BEF = 180°

⟹ ∠CBF + ∠BEF= 90° ..... (1)

Now, in quadrilateral AFOE, Z∠AFO + Z∠AEO = 180°. [∴ BE ⊥ AC and CF ⊥ AB]

But, we know that if the sum of any pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic. Therefore, AFOE is a cyclic quadrilateral.

∴ ∠OEF = ∠OAF [Angles in the same segment] ...(2)

∵ ∠CBF + ∠BEF = 90° [From (1)]

∴ ∠DBA + ∠OEF = 90°

⟹ ∠DBA + ∠DAB = 90° [From (2)] ..(3)

But, ∠DBA + ∠DAB + ∠ADB = 180°

⟹ 90° + ∠ADB = 180° [From (3)]

∴ ∠ADB = 90°

Hence, AD ⊥ BC.

(Q.E.D.)