Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.
Answers
Step-by-step explanation:
Given: In ΔABC, ∠B = 90°
D is midpoint of AC
To Prove:
Construction: Draw a line l passing through D and parallel to AB
Proof: Since l || AB
∠ABC = ∠DEC = 90° → (1)
In ΔABC, D is midpoint of AC
DE || AB
∴ E is midpoint of BC (By converse of Midpoint theorem)
⇒ BE = EC → (2)
In ΔDEB and ΔDEC
BE = EC [From (2)]
∠DEB = ∠DEC = 90º [from (1)]
DE is common side
ΔDEB ≅ ΔDEC (By SAS property) ⇒DC = BD
But DC is half AC [D is midpoint of AC]
⇒ BD is Half of AC
Answer:
Step-by-step explanation:
Let P be the mid point of the hypotenuse of the right △ABC right angled at B
Draw a line parallel to BC from P meeting B at O
Join PB
In △PAD and △PBD
∠PDA=∠PDB=90
∘
each due to conv of mid point theorem
PD=PD (common)
AD=DB (As D is mid point of AB)
So △ PAD and PBD are congruent by SAS rule
PA=PB (C.P.C.T)
As PA=PC (Given as P is mid-point)
∴PA=PC=PB