Exercise 9.3 Part 2
Question 5: D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
- BDFE is a parallelogram
- ar(DEF) = ¼ ar(ABC)
- ar(BDEF) = ½ ar(ABC)
From mid-point theorem; BD||EF
BD = ½ BC (Because D is midpoint)
Hence, EF = BD
As EF = BD
So, BDFE is a parallelogram
Similarly, it can be proved that EFDC and AEDF are parallelograms.
As BD = CD = EF
Hence, ar(BDFE) = ar(EFDC)
In triangles BED and EFD;
BD = EF
DE = DE
So, from SSS theorem; ΔBDE ≈ ΔEFD
Similarly, it can be proven that ΔEFD ≈ ΔCDF
Similarly, it can be proven that ΔEFD ≈ ΔFEA
Thus, ΔBDE ≈ ΔEFD ≈ ΔCDF ≈ ΔFEA
Hence, ar(DEF) = ¼ ar(ABC)
As parallelogram BDFE is composed of two triangles
Hence, ar(BDFE) = ½ ar(ABC) proved.
Question 6: In the given figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that
- ar(DOC) = ar(AOB)
- ar(DCB) = ar(ACB)
- DA||CB or ABCD is a parallelogram
(Hint: From D and B, draw perpendiculars to AC. )
Answer: (i) In triangles DOC and AOB;
DC = AB (given)
DO = BO (given)
Angle DOC = Angle AOB (Vertically opposite angles)
Hence, from SAS theorem; ΔDOC ≈ ΔAOB
Hence, ar(DOC) = ar(AOB)
(ii) In triangles DCB and ACB;
DC = AB (Given)
CB = CB (Common side)
Hence, from SSS theorem; ΔDCB ≈ ΔACB
Hence, ar(DCB) = ar(ACB) Proved
(iii) As opposite sides are equal hence, the quadrilateral ABCD is a parallelogram and DA||CB is proved.
Question 7: D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC) = ar(EBC). Prove that DE||BC.
Since ar(DBC) = ar(EBC)
And these triangles have a common base, i.e. BC
Hence, they must be having the same altitude.
So, they are between same parallels.
Hence, DE||BC proved.
Question 8: XY is a line parallel to side BC of a triangle ABC. If BE||AC and CF||AB meet XY at E and F respectively, show that ar(ABE) = ar(ACF)
Answer: BEYC is a parallelogram because EB||YC (Given EB||AC) and EY||BC (because XY ||BC)
Ex 9.1 and 9.2
Ex 9.3 Part 1
Ex 9.3 Part 2
Ex 9.3 Part 3
Ex 9.3 Part 4
In triangle AEB and parallelogram BEYC;
ar(AEB) = ½ ar(BEYC) (because triangle and parallelogram are between same parallels.)
Similarly, ar(ACF) = ½ ar(BXFC) (because triangle and parallelogram are between same parallels).
Now, ar(BEYC) = ar(BXFC) (because they are between same parallels)
Hence, ar(AEB) = ar(ACF) proved