Question

in triangle abc altitude a d bisects BC prove that triangle ADB is congruent to triangle ADC write equal pair of sides of these two angles​

Answer 1

Answer:

Given a triangle ABC with acute angles, let A*, B*, C* be the feet of the altitudes of the triangle: A*, B*, C* are points on the sides of the triangle so that AA* BB*, CC* are altitudes.

Then we have proved earlier that the altitudes are concurrent at a point H.  (The proof used the relationship between the perpendicular bisectors of the sides of a triangle and the altitudes of its midpoint triangle).

The orthic triangle of ABC is defined to be A*B*C*.  This triangle has some remarkable properties that we shall prove:

The altitudes and sides of ABC are interior and exterior angle bisectors of orthic triangle A*B*C*, so H is the incenter of A*B*C* and A, B, C are the 3 ecenters (centers of escribed circles).

The sides of the orthic triangle form an "optical" or "billiard" path reflecting off the sides of ABC.

From this it can be proved that the orthic triangle A*B*C* has the smallest perimeter of any triangle with vertices on the sides of ABC.

Step-by-step explanation:

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