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
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
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that:
(i) ΔDCE ≅ ΔLBE
(ii) AB = BL.
(iii) AL = 2DC
In the following figure, BL = CM.
Prove that AD is a median of triangle ABC.
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.
Prove that: ΔBEC ≅ ΔDCF.
In the figure, given below, triangle ABC is right-angled at B. ABPQ and ACRS are squares. 
Prove that:
(i) ΔACQ and ΔASB are congruent.
(ii) CQ = BS.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
Which of the following is not a criterion for congruence of triangles?
