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 Fig. 7.23). Show that:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
iii) ΔDBC ≅ ΔACB
iv) CM = 1/2 AB
Solution:
Given: M is the mid-point of hypotenuse AB, ∠C = 90° and DM = CM
To Prove:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
iii) ΔDBC ≅ ΔACB
iv) CM = 1/2 AB
i) In ΔAMC and ΔBMD,
AM = BM (M is the mid - point of AB)
∠AMC = ∠BMD (Vertically opposite angles)
CM = DM (Given)
∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)
∴ AC = BD (By CPCT)
Also, ∠ACM = ∠BDM (By CPCT)
ii) ∠DBC is a right angle.
We know that, ∠ACM = ∠BDM (proved above)
But, ∠ACM and ∠BDM are alternate interior angles. Since alternate angles are equal, it can be said that DB || AC.
∠DBC + ∠ACB = 180° (Co-interior angles)
∠DBC + 90° = 180° [Since, ΔACB is a right angled triangle]
∴ ∠DBC = 90°
Thus, ∠DBC is a right angle.
iii) In ΔDBC and ΔACB,
DB = AC (Already proved)
∠DBC = ∠ACB = 90° (Proved above)
BC = CB(Common)
∴ Δ DBC ≅ Δ ACB (SAS congruence rule)
iv) CM = 1/2 AB
Since Δ DBC ≅ Δ ACB
AB = DC (By CPCT)
⇒ 1/2 AB = 1/2 DC
It is given that M is the midpoint of DC and AB.
CM = 1/2 DC = 1/2 AB
∴ CM = 1/2 AB
☛ Check: NCERT Solutions Class 9 Maths Chapter 7
Video Solution:
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 Fig. 7.23). Show that:
i) ΔAMC ≅ ΔBMD
ii) ∠DBC is a right angle.
iii) ΔDBC ≅ ΔACB
iv) CM = 1/2 AB
NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.1 Question 8
Summary:
If 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 and if point D is joined to point B, then it is proved that ΔAMC ≅ ΔBMD using SAS congruence, ∠DBC is a right angle, ΔDBC ≅ ΔACB using SAS congruence, and CM = 1/2AB.
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