Advertisements
Advertisements
प्रश्न
In the given figure: AB//FD, AC//GE and BD = CE;
prove that:
- BG = DF
- CF = EG

Advertisements
उत्तर
In the given figure AB || FD,
⇒ ∠ABC =∠FDC
Also AC || GE,
⇒ ∠ACB = ∠GEB
Consider the two triangles ΔGBE and ΔFDC
∠B = ∠D ...(Corresponding angle)
∠C = ∠E ...(Corresponding angle)
Also given that
BD = CE
⇒ BD + DE = CE + DE
⇒ BE = DC
∴ By Angle-Side-Side-Angle criterion of congruence
ΔGBE ≅ ΔFDC
∴ `"GB"/"FD" = "BE"/"DC" = "GE"/"FC"`
But BE = DC
⇒ `"BE"/"DC" = "BE"/"BE"` = 1
∴ `"GB"/"FD" = "BE"/"DC"` = 1
⇒ GB = FD
∴ `"GE"/"FC" = "BE"/"DC"` = 1
⇒ GE = FC
APPEARS IN
संबंधित प्रश्न
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.

You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

ABCD is a square, X and Yare points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
In triangles ABC and CDE, if AC = CE, BC = CD, ∠A = 60°, ∠C = 30° and ∠D = 90°. Are two triangles congruent?
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
D, E, F are the mid-point of the sides BC, CA and AB respectively of ΔABC. Then ΔDEF is congruent to triangle
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD.
Prove that :
(i) ΔABD and ΔECD are congruent.
(ii) AB = CE.
(iii) AB is parallel to EC
In a ΔABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that: AE is parallel to BC.
