Question

5. ABC is an isosceles triangle in which altitudes BE and CF are drawn to

equal sides AC and AB respectively. Then:

a. BE>CF

b. BE<CF

c. BE=CF

d. None of the above​

Answer 1

c)BE=CF

Hope it helps you

Answer 2

Answer:

and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ≌ ΔCDA.

Ans. ∵l || m and AC is a transversal.

∴ ∠BAC = ∠DCA

[Alternate interior angles]

Also p || q and AC is a transversal,

∴ ∠BCA = ∠DAC

[Alternate interior angles]

Now, in ΔABC and ΔCDA,

∠BAC = ∠DCA

[Proved]

∠BCA = ∠DAC

[Proved]

CA = AC

[Common]

∴ Using ASA criteria, we have ΔABC ≌ ΔCDA

Step-by-step explanation:

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