Question

in figure Angle B is a right angle in triangle abc and D is the midpoint of AC also d e parallel to a b and c intersects BC at P show that P is the midpoint of BC and de perpendicular to bc and bd=ad​

Answer 1

Answer:

Lets write the known facts first

ABC is a Right Angled Triangle, Right - Angled at B

D is Mid-Point of AC, DE is Parallel to AB, intersecting BC at P

Now, As DP is parallel to AB

Angle B = Angle P (corresponding angles)

Angle C is common,

So by sum of all angles of a triangle, Angle D = Angle A

Therefore, Triangle ABC is similar to Triangle DPC.

As DC = 1/2 AC,

Using similarity of these Triangles we can say PC = 1/2 BC

Therefore, DP is a Perpendicular Bisector of BC.

Now in Triangles DEC and DEB

DP= DP

PC = BP (Previously Proved)

Angle DPC = Angle DPB

So, we can say that Triangle DPC is Congruent to DPB

Therefore, BD = CD

And CD = AD (Given)

So, BD = AD

Answer 2

Answer:

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Step-by-step explanation:

Lets write the known facts first

ABC is a Right Angled Triangle, Right - Angled at B

D is Mid-Point of AC, DE is Parallel to AB, intersecting BC at P

Now, As DP is parallel to AB

Angle B = Angle P (corresponding angles)

Angle C is common,

So by sum of all angles of a triangle, Angle D = Angle A

Therefore, Triangle ABC is similar to Triangle DPC.

As DC = 1/2 AC,

Using similarity of these Triangles we can say PC = 1/2 BC

Therefore, DP is a Perpendicular Bisector of BC.

Now in Triangles DEC and DEB

DP= DP

PC = BP (Previously Proved)

Angle DPC = Angle DPB

So, we can say that Triangle DPC is Congruent to DPB

Therefore, BD = CD

And CD = AD (Given)

So, BD = AD