Ex7.1, 8
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
ΔAMC ≅ ΔBMD
Given:
∠ ACB = 90°
M is the mid-point of AB
So, AM = BM
Also, DM = CM
To prove: ΔAMC ≅ ΔBMD
Proof:
Lines CD & AB intersect…
So, ∠AMC = ∠BMD
In ΔAMC and ΔBMD,
AM = BM
∠AMC = ∠BMD
CM = DM
∴ ΔAMC ≅ ΔBMD
Ex7.1, 8
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(ii) ∠DBC is a right angle.
From part 1,
ΔAMC ≅ ΔBMD
∴ ∠ACM = ∠BDM
But ∠ACM and ∠BDM are alternate interior angles
for lines AC & BD
If a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel.
So, BD || AC
Now, Since
DB || AC and Considering BC as transversal
, ⇒ ∠DBC + ∠ACB = 180°
⇒ ∠DBC + 90° = 180°
⇒ ∠DBC = 180° – 90°
⇒ ∠DBC = 90°
Hence, ∠ DBC is a right angle
Hence proved
Ex 7.1, 8
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(iii) ΔDBC ≅ ΔACB
From part 1,
ΔAMC ≅ ΔBMD
AC = BD
In ΔDBC and ΔACB,
DB = AC
∠DBC = ∠ACB
BC = CB
∴ ΔDBC ≅ ΔACB
Ex7.1, 8
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
(iv) CM = 1/2 AB
From part 3,
ΔDBC ≅ ΔACB
∴ DC = AB
1/2 DC = 1/2 AB
CM = 1/2 AB
Hence proved

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.