Advertisements
Advertisements
Question
In a right-angled triangle ABC. ∠ABC = 90° and D is the midpoint of AC. Prove that BD = `(1)/(2)"AC"`.
Advertisements
Solution
Draw line segment DE || CB, which meets AB at point E.
Now, DE || CB and AB is the transversal,
∴ ∠AED = ∠ABC ....(corrresponding angles)
∠ABC = 90° ....(given)
⇒ ∠AED = 90°
Also, as D is the mid-point of AC and DE || CB,
DE bisects side AB,
I.e. AE = BE ....(i)
In ΔAED and ΔBED,
∠AED = ∠BED ....(Each 90°)
AE = BE ....[From (i)]
DE = DE ....(Common)
∴ ΔAED ≅ ΔBEd ....(By SAS Test)
⇒ AD = BD ....(C.P.C.T.C)
⇒ BD = AC
⇒ BD = `(1)/(2)"AC"`.
APPEARS IN
RELATED QUESTIONS
ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
In the below Fig, ABCD and PQRC are rectangles and Q is the mid-point of Prove thaT
i) DP = PC (ii) PR = `1/2` AC
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.
Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:
(i) AQ // BS
(ii) DS = 3 Rs.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In ΔABC, D, E and F are the midpoints of AB, BC and AC.
If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: ΔGEA ≅ ΔGFD
D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.
