triangle ABC is a right angle triangle at b and p is the midpoint of AC and PQ is parallel to BC prove that PQ is perpendicular to ab and q is the midpoint of AB
Answers
Answer:
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QR
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof:
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem)
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem)
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem) In ∴ ABP, Q is the mid point of AB and QR||BP.
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem) In ∴ ABP, Q is the mid point of AB and QR||BP.∴ R is the mid point of AP. (Converse of mid point theorem)
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem) In ∴ ABP, Q is the mid point of AB and QR||BP.∴ R is the mid point of AP. (Converse of mid point theorem)⇒ AP = 2AR
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem) In ∴ ABP, Q is the mid point of AB and QR||BP.∴ R is the mid point of AP. (Converse of mid point theorem)⇒ AP = 2AR
Given: In ∆ABC, P is the mid point of BC. PQ||CA, PQ meets AB in Q. QR||BC, QR meets AP in R.To prove: 1. AP = 2 AR 2. BC = 4 QRProof: In ∆ABC, P is the mid point of BC and PQ||AB.∴ Q is the mid point of AB (Converse of mid-point theorem) In ∴ ABP, Q is the mid point of AB and QR||BP.∴ R is the mid point of AP. (Converse of mid point theorem)⇒ AP = 2AR In ∆ABP, Q is the mid point of AB and R is the mid point of AP.
Answer:
In ΔAMD And ΔABC
∠A is common;
∠B=∠M (corresponding angles)
∠C=∠D (BC is parallel to MD and BC is perpendicular to AC therefore, MD is also perpendicular to AC)..........(i)
ΔAMD≅ΔABC (by AAA)
∴
AB
AM
=
AC
AD
=
2
1
⇒
AD+CD
AD
=
2
1
⇒2AD=AD+CD
⇒AD=CD
Hence the D is mid point of AC